find the solutions to z squared = 1-i square root of 3

To find the solutions to the equation \( z^2 = 1 - i\sqrt{3} \), let's break it down step by step.

### Step 1: Express the complex number in polar form
The given complex number is \( 1 - i\sqrt{3} \). We need to express this in polar form, which is \( r \text{cis} \theta = r(\cos \theta + i \sin \theta) \), where \( r \) is the modulus and \( \theta \) is the argument.

1. **Modulus** \( r \):
   \[
   r = \sqrt{(1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2
   \]

2. **Argument** \( \theta \):
   The argument \( \theta \) is the angle the complex number makes with the positive real axis. Since the complex number is \( 1 - i\sqrt{3} \), it lies in the fourth quadrant.

   \[
   \theta = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}
   \]

So, in polar form, \( 1 - i\sqrt{3} = 2 \text{cis}\left(-\frac{\pi}{3}\right) \).

### Step 2: Use De Moivre's Theorem
We know that \( z^2 = r \text{cis} \theta \), and we are trying to find \( z \). So, to solve for \( z \), we need to take the square root of both sides.

Let \( z = r^{1/2} \text{cis} \left(\frac{\theta + 2k\pi}{2}\right) \), where \( k = 0, 1 \) gives the two distinct solutions.

1. **Modulus**: 
   \[
   \sqrt{2} = \sqrt{r}
   \]

2. **Arguments**:
   For \( k = 0 \):
   \[
   \theta_1 = \frac{-\frac{\pi}{3} + 2(0)\pi}{2} = -\frac{\pi}{6}
   \]

   For \( k = 1 \):
   \[
   \theta_2 = \frac{-\frac{\pi}{3} + 2(1)\pi}{2} = \frac{5\pi}{6}
   \]

### Step 3: Write the solutions
The two solutions are:

1. \( z_1 = \sqrt{2} \text{cis} \left(-\frac{\pi}{6}\right) = \sqrt{2} \left( \cos \left(-\frac{\pi}{6}\right) + i \sin \left(-\frac{\pi}{6}\right) \right) \)
   \[
   z_1 = \sqrt{2} \left( \frac{\sqrt{3}}{2} - i \frac{1}{2} \right) = \frac{\sqrt{6}}{2} - i \frac{\sqrt{2}}{2}
   \]

2. \( z_2 = \sqrt{2} \text{cis} \left(\frac{5\pi}{6}\right) = \sqrt{2} \left( \cos \left(\frac{5\pi}{6}\right) + i \sin \left(\frac{5\pi}{6}\right) \right) \)
   \[
   z_2 = \sqrt{2} \left( -\frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = -\frac{\sqrt{6}}{2} + i \frac{\sqrt{2}}{2}
   \]

### Final Solutions:
\[
z_1 = \frac{\sqrt{6}}{2} - i \frac{\sqrt{2}}{2}, \quad z_2 = -\frac{\sqrt{6}}{2} + i \frac{\sqrt{2}}{2}
\]

Would you like more details on any step or have questions?

Here are 5 related questions to expand on this topic:
1. How do you convert any complex number into polar form?
2. What is De Moivre’s theorem and how is it used for solving powers of complex numbers?
3. What are the geometric interpretations of complex number solutions?
4. Can this method be generalized for solving cubic equations with complex numbers?
5. How do you solve equations involving higher powers of complex numbers?

**Tip**: Always check in which quadrant the complex number lies when finding the argument (angle), as it affects the sign of the trigonometric functions.

find the solutions to z squared = 1 - i square root of 3

Discover how to solve the equation z^2 = 1 - i√3 by converting the complex number into polar form and applying De Moivre's Theorem. Follow a detailed step-by-step solution suitable for advanced high school and undergraduate math students.

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